# Cardinality of a set

• Apr 29th 2010, 12:26 PM
posix_memalign
Cardinality of a set
How can I show that the set {$\displaystyle (A, B) \in P(S) \times P(S) | A \subseteq B$} contains $\displaystyle 3^n$ elements, where S is a set of n elements and P(S) is the power set of S.

[SOLVED]
• Apr 29th 2010, 01:30 PM
Plato
Quote:

Originally Posted by posix_memalign
Let S be a set of n elements and consider the power set P(S) of S.
How can I show that the set {$\displaystyle (A, B) \in P(S) \times P(S) | A \subseteq B$} contains $\displaystyle 3^n$ elements?

Here is some notation. $\displaystyle |A|$ is the number of elements is set $\displaystyle A$.
Thus $\displaystyle |S|=n$ and $\displaystyle |\mathcal{P}(S)|=2^n$.

For each set $\displaystyle A$ that are $\displaystyle 2^{n-|A|}$ subsets of $\displaystyle S$ that are super-sets of $\displaystyle A$
i.e. $\displaystyle |\{X\in \mathcal{P}(S):A\subseteq X\}|=2^{n-|A|}$
Hence that is the number of wanted pairs $\displaystyle (A,B)$.

To count them find the sum $\displaystyle \sum\limits_{k = 0}^n {\binom{n}{k} \cdot 2^{n - k} }$. Do you see why?

Now use the binomial expansion theorem on $\displaystyle 3^n=(1+2)^n$
• Apr 29th 2010, 03:44 PM
posix_memalign
Quote:

Originally Posted by Plato
To count them find the sum $\displaystyle \sum\limits_{k = 0}^n {\binom{n}{k} \cdot 2^{n - k} }$. Do you see why?

After you have arrived at this then it is obvious why it is $\displaystyle 3^n$ but I don't see how you arrive at this statement unfortunately no.
• Apr 29th 2010, 04:03 PM
Plato
Quote:

Originally Posted by posix_memalign
After you have arrived at this then it is obvious why it is $\displaystyle 3^n$ but I don't see how you arrive at this statement unfortunately no.

Then I must ask you if you understand that the number of subsets having
$\displaystyle A$ as a subset is $\displaystyle 2^{n-|A|}?$
If you do then there are $\displaystyle \sum\limits_{k = 0}^n {\binom{n}{k}}$ subsets of each ‘size’.